Question 878776
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Hi
Below You will find a  similar workup
Particulars for this problem
p(error) = .10,  n = 8337, mean = 833.7 and SD = sqrt(833.7*.90) = 27.392
Use 834.5  and 833.5 as shown below
Stattrek.com gives a).448  b) .512  c) .015
doing the work will give You the 4 decimal points
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I.  recommend using a TI Calculator...
II. Must Learn when to recognize a Binomial Distribution, when You see one.
Most recognizable when You see a % of a certain event given.  Generally when samples are very Large(>1000), 
one uses a <u>normal approximation</u> to the binomial.
Trick is: When Using the normal <u>approximation</u>, one uses 'endpoints' (using .5)

a)p(error) = .10,  n = 5370, mean = 537 and SD = sqrt(537*.90) = 21.98
{{{z[530.5] = (530.5-537)/21.98 = -.2957}}}, {{{z[529.5] = (529.5-537)/21.98 = -.3412}}}, 
P(x > 530) = 1 - P(z &#8804; -.2957) = 1 - .3837= .6163
b) P(x &#8804; 530) = P(z &#8804; -.2957) = .3837
c) P(x = 530) = P(z &#8804; -.2957) - P(z &#8804; -.3412) =.3837 - .3665 = .0172
Again, while it is not to 4 decimal points.. one can verify this using stattrek.com (Binomial distributions)
If Using TI...Using syntax: normalcdf(smaller, larger, µ, &#963;).
normalcdf(-9999, 530.5, 537, 21.98) and normal(-9999,529.5,537, 21.98)
would have made short work of this
Note: The -9999 is used as the smaller value to be at least 5 standard deviations from the mean.